I am practicing creating a randomly generating integers in an array, then randomizing the elements in the array. All is well when I print the numbers, but there seems to be one element that does not print when I am displaying the randomized elements. Is there a step I am leaving out?
public class shufflingArrays {
public static void main(String[] args) {
int[] myList = new int[10];
System.out.println("Numbers:");
for(int i = 0; i < myList.length; i++) {
myList[i] = (int)(Math.random() * 100);
System.out.print(myList[i] + " ");
}
System.out.println("\nRandomized:");
for (int i = myList.length - 1; i > 0; i--){
//Generate index j randomly with 0 <= j <= i
int j = (int)(Math.random() * (i + 1));
//Swap myList[i]; with myList[j]
int temp = myList[i];
myList[i] = myList[j];
myList[j] = temp;
System.out.print(myList[i] + " ");
}
}
Your for loop has condition i > 0, which means when i == 0 it will terminate and not print out the first array element.
However, if you're doing the Fisher-Yates shuffle, as it appears, you do indeed need to go from myList.length-1 to 1, so your initial code was correct. You then can't print out all the elements in the array from the same loop, so either use another loop after to print out the elements, or add System.out.print(myList[0]); after.
Ex: for (int i = 4; i > 0; i--)
will run the for loop when i = 4, 3, 2, 1 only and not when i = 0, because the condition there is i > 0. Change the i > 0 condition in for (int i = myList.length - 1; i > 0; i--) to i >= 0 and you will get what you want.
Related
New to java and am trying to create a program/method that will take an int array, and return another int array but it replaces the values of the indexes with the value of the elements. (Example {2,1,3} will return {0,0,1,2,2,2}
public static void main(String[] args) {
int[] pracArray = {2, 1, 3};
int sum = 0;
for (int i = 0; i < pracArray.length; i++)
{
sum = sum + pracArray[i];
}
System.out.println("Amount of array indexes: " + sum);
int[] newArray = new int[sum];
System.out.println(Arrays.toString(pracArray));
for (int i = 0; i < pracArray.length; i++)
{
for (int j = 0; j < pracArray[i]; j++)
{
newArray[j] = i;
}
}
System.out.println(Arrays.toString(newArray));
}
}
Currently I am getting [2,2,2,0,0,0]. I have tried changing the how many times each for loop iterates with no avail. I have also tried to make the elements of newArray equal to a counter ( int count = 0; and having count++ in the for loop) since the values of the new array will always be 0 - however many runs.
Given the length of your array is 3, your outer 'i' loop is iterating through the values 0,1,2. That means your inner 'j' loop never writes to index 3,4,5 (hence why they are 0 in the output), and why the first 3 indexes are set to '2' (2 is the last indexed processed in the 'i' loop). Try this instead...
int h = 0;
for (int i = 0; i < pracArray.length; i++)
{
for (int j = 0; j < pracArray[i]; j++)
{
newArray[h] = i;
h++;
}
}
I need to take 2 inputs from user size of array and then elements of that same array.
I need to print every third element of array.
Example Array: 1,2,3,4,5,6,7,8,9
Desired output: 3,6,9
Getting: 9,6,3
class Demo {
public static void main(String[] args) {
int x;
int[] y;
Scanner tastatura = new Scanner(System.in);
System.out.println("Enter size of array:");
x = tastatura.nextInt();
y = new int[x];
System.out.println("Enter the elements of array:");
for (int i = 0; i < x; i++) {
y[i] = tastatura.nextInt();
}
System.out.println("\n Every third element of array is : ");
for (int i = y.length - 1; i >= 0; i = i - 3) {
System.out.println(y[i]);
}
tastatura.close();
}
You were close! You just have the iteration order reversed!
for (int i = y.length - 1; i >= 0; i = i - 3) {
System.out.println(y[i]);
}
Should be:
for (int i = 2; i < y.length; i += 3) {
System.out.println(y[i]);
}
Take note that this can throw an ArrayIndexOutOfBoundsException if your array does not contain at least 3 elements, so you should handle that somewhere.
use the modulus operator to find each 3rd item.
example
0 % 3 = 0
1 % 3 = 1
2 % 3 = 2
3 % 3 = 0
for (int i =0 ; i < y.length - 1; i = i + 3) {
System.out.println(y[i]);
}
The way you are printing is in reverse order .. Correct that
You are on the right track I would strongly recommend using a modulus:
for (int i = 0; i < y.length; i++) {
if(i%3==0){
System.out.println(y[i]);
}//if statement
}//for loop
I am trying to sort numbers into two categories. Number (actual number) and count (how many occurrences of this number), all of the numbers are stored in a array of 50 integers. I sort this array in descending order using a bubble sort method.
My print method needs work. The code compiles just fine but when I run the code nothing is output.
Why is my code not printing anything?
Here is my code
public class HW5{
public static void main(String[] args) {
int[] array = new int[50];
bubbleSort(array, 'D');
printArray(array);
}
public static void bubbleSort(int[] array, char d){
int r = (d=='D') ? -1 : 1 ;
for (int f = 0; f < array.length - 1; f ++){
for (int index = 0; index < array.length - 1; index++){
if (array[index] > array[index + 1]){
}
}
}
}
public static void printArray(int[] array){
int count = 0;
int i = 0;
for(i = 0; i < array.length - 1; i++){
if (array[i]== array[i + 1]){
count = count + 1;
}else{
System.out.printf(count + "/t" + array[i]);
count = 0;
}
}
}
}
Why is my code not printing anything?
object array is holding 50 elements all of them with the value set to zero and the printArray method will print if and only IF this condition is false
array[i] != array[i + 1]
but since all elements in the array are 0... you just dont print anything...
Your code is printing anything because you aren't setting your array values to anything. Because you aren't setting your array to anything, java defaults all the values to 0. Your code doesn't output if array[i] == array[i+1] I'd change your print method to this:
public static void printArray(int[] array){
int count = 0;
int i = 0;
for(i = 0; i < array.length - 1; i++){
if (array[i]== array[i + 1]){
count = count + 1;
}else{
System.out.print(count);
count = 0;
}
System.out.print(array[i]); //Moved this line out of the if/else statement so it will always print the array at i
}
}
I only changed the line I commented on. However, if you did change the values of your array, your original code would work. For random values, first you need to import java.util.Math then do the following:
for(int i = 0; i < array.length; i++)
array[i] = (int)Math.random() * 100; //Sets the array at i to a random number between 0 and 100 (non-inclusive)
This will help your code above to work as you wanted. Hope this helps!
EDIT: Fixed a grammar error.
This question already has answers here:
How do I reverse an int array in Java?
(47 answers)
Closed 8 years ago.
I have an array of n elements and these methods:
last() return the last int of the array
first() return the first int of the array
size() return the length of the array
replaceFirst(num) that add the int at the beginning and returns its position
remove(pos) that delete the int at the pos
I have to create a new method that gives me the array at the reverse order.
I need to use those method. Now, I can't understand why my method doesn't work.
so
for (int i = 1; i
The remove will remove the element at the position i, and return the number that it is in that position, and then with replaceFirst will move the number (returned by remove) of the array.
I made a try with a simple array with {2,4,6,8,10,12}
My output is: 12 12 12 8 6 10
so if I have an array with 1,2,3,4,5
for i = 1; I'm gonna have : 2,1,3,4,5
for i=2 >3,2,1,4,5
etc
But it doesn't seem to work.
Well, I'll give you hints. There are multiple ways to reverse an array.
The simplest and the most obvious way would be to loop through the array in the reverse order and assign the values to another array in the right order.
The previous method would require you to use an extra array, and if you do not want to do that, you could have two indices in a for loop, one from the first and next from the last and start swapping the values at those indices.
Your method also works, but since you insert the values into the front of the array, its going to be a bit more complex.
There is also a Collections.reverse method in the Collections class to reverse arrays of objects. You can read about it in this post
Here is an code that was put up on Stackoverflow by #unholysampler. You might want to start there: Java array order reversing
public static void reverse(int[] a)
{
int l = a.length;
for (int j = 0; j < l / 2; j++)
{
int temp = a[j]
a[j] = a[l - j - 1];
a[l - j - 1] = temp;
}
}
int[] reverse(int[] a) {
int len = a.length;
int[] result = new int[len];
for (int i = len; i > 0 ; i--)
result[len-i] = a[i-1];
return result;
}
for(int i = array.length; i >= 0; i--){
System.out.printf("%d\n",array[i]);
}
Try this.
If it is a Java array and not a complex type, the easiest and safest way is to use a library, e.g. Apache commons: ArrayUtils.reverse(array);
In Java for a random Array:
public static void reverse(){
int[] a = new int[4];
a[0] = 3;
a[1] = 2;
a[2] = 5;
a[3] = 1;
LinkedList<Integer> b = new LinkedList<Integer>();
for(int i = a.length-1; i >= 0; i--){
b.add(a[i]);
}
for(int i=0; i<b.size(); i++){
a[i] = b.get(i);
System.out.print(a[i] + ",");
}
}
Hope this helps.
Reversing an array is a relatively simple process. Let's start with thinking how you print an array normally.
int[] numbers = {1,2,3,4,5,6};
for(int x = 0; x < numbers.length; x++)
{
System.out.println(numbers[x]);
}
What does this do? Well it increments x while it is less than numbers.length, so what is actually happening is..
First run : X = 0
System.out.println(numbers[x]);
// Which is equivalent to..
System.out.println(numbers[0]);
// Which resolves to..
System.out.println(1);
Second Run : X = 1
System.out.println(numbers[x]);
// Which is equivalent to..
System.out.println(numbers[1]);
// Which resolves to..
System.out.println(2);
What you need to do is start with numbers.length - 1, and go back down to 0. To do this, you need to restructure your for loop, to match the following pseudocode..
for(x := numbers.length to 0) {
print numbers[x]
}
Now you've worked out how to print, it's time to move onto reversing the array. Using your for loop, you can cycle through each value in the array from start to finish. You'll also be needing a new array.
int[] revNumbers = new int[numbers.length];
for(int x = numbers.length - 1 to 0) {
revNumbers[(numbers.length - 1) - x] = numbers[x];
}
int[] noArray = {1,2,3,4,5,6};
int lenght = noArray.length - 1;
for(int x = lenght ; x >= 0; x--)
{
System.out.println(noArray[x]);
}
}
int[] numbers = {1,2,3,4,5};
int[] ReverseNumbers = new int[numbers.Length];
for(int a=0; a<numbers.Length; a++)
{
ReverseNumbers[a] = numbers.Length - a;
}
for(int a=0; a<ReverseNumbers.Length; a++)
Console.Write(" " + ReverseNumbers[a]);
int[] numbers = { 1, 2, 3, 4, 5, 6 };
reverse(numbers, 1); >2,1,3,4,5
reverse(numbers, 2); >3,2,1,4,5
public int[] reverse(int[] numbers, int value) {
int index = 0;
for (int i = 0; i < numbers.length; i++) {
int j = numbers[i];
if (j == value) {
index = i;
break;
}
}
int i = 0;
int[] result = new int[numbers.length];
int forIndex = index + 1;
for (int x = index + 2; x > 0; x--) {
result[i] = numbers[forIndex--];
++i;
}
for (int x = index + 2; x < numbers.length; x++) {
result[i] = numbers[x];
++i;
}
return result;
}
OK, so I found this question from a few days ago but it's on hold and it won't let me post anything on it.
***Note: The values or order in the array are completely random. They should also be able to be negative.
Someone recommended this code and was thumbed up for it, but I don't see how this can solve the problem. If one of the least occurring elements isn't at the BEGINNING of the array then this does not work. This is because the maxCount will be equal to array.length and the results array will ALWAYS take the first element in the code written below.
What ways are there to combat this, using simple java such as below? No hash-maps and whatnot. I've been thinking about it for a while but can't really come up with anything. Maybe using a double array to store the count of a certain number? How would you solve this? Any guidance?
public static void main(String[] args)
{
int[] array = { 1, 2, 3, 3, 2, 2, 4, 4, 5, 4 };
int count = 0;
int maxCount = 10;
int[] results = new int[array.length];
int k = 0; // To keep index in 'results'
// Initializing 'results', so when printing, elements that -1 are not part of the result
// If your array also contains negative numbers, change '-1' to another more appropriate
for (int i = 0; i < results.length; i++) {
results[i] = -1;
}
for (int i = 0; i < array.length; i++) {
for (int j = 0; j < array.length; j++) {
if (array[j] == array[i]) {
count++;
}
}
if (count <= maxCount) { // <= so it admits number with the SAME number of occurrences
maxCount = count;
results[k++] = array[i]; // Add to 'results' and increase counter 'k'
}
count = 0; // Reset 'count'
}
// Printing result
for (int i : results) {
if (i != -1) {
System.out.println("Element: " + i + ", Number of occurences: " + maxCount);
}
}
}
credit to: https://stackoverflow.com/users/2670792/christian
for the code
I can't thumbs up so I'd just like to say here THANKS EVERYONE WHO ANSWERED.
You can also use an oriented object approach.
First create a class Pair :
class Pair {
int val;
int occ;
public Pair(int val){
this.val = val;
this.occ = 1;
}
public void increaseOcc(){
occ++;
}
#Override
public String toString(){
return this.val+"-"+this.occ;
}
}
Now here's the main:
public static void main(String[] args) {
int[] array = { 1,1, 2, 3, 3, 2, 2, 6, 4, 4, 4 ,0};
Arrays.sort(array);
int currentMin = Integer.MAX_VALUE;
int index = 0;
Pair[] minOcc = new Pair[array.length];
minOcc[index] = new Pair(array[0]);
for(int i = 1; i < array.length; i++){
if(array[i-1] == array[i]){
minOcc[index].increaseOcc();
} else {
currentMin = currentMin > minOcc[index].occ ? minOcc[index].occ : currentMin;
minOcc[++index] = new Pair(array[i]);
}
}
for(Pair p : minOcc){
if(p != null && p.occ == currentMin){
System.out.println(p);
}
}
}
Which outputs:
0-1
6-1
Explanation:
First you sort the array of values. Now you iterate through it.
While the current value is equals to the previous, you increment the number of occurences for this value. Otherwise it means that the current value is different. So in this case you create a new Pair with the new value and one occurence.
During the iteration you will keep track of the minimum number of occurences you seen.
Now you can iterate through your array of Pair and check if for each Pair, it's occurence value is equals to the minimum number of occurences you found.
This algorithm runs in O(nlogn) (due to Arrays.sort) instead of O(n²) for your previous version.
This algorithm is recording the values having the least number of occurrences so far (as it's processing) and then printing all of them alongside the value of maxCount (which is the count for the value having the overall smallest number of occurrences).
A quick fix is to record the count for each position and then only print those whose count is equal to the maxCount (which I've renamed minCount):
public static void main(String[] args) {
int[] array = { 5, 1, 2, 2, -1, 1, 5, 4 };
int[] results = new int[array.length];
int minCount = Integer.MAX_VALUE;
for (int i = 0; i < array.length; i++) {
for (int j = 0; j < array.length; j++) {
if (array[j] == array[i]) {
results[i]++;
}
}
if (results[i] <= minCount) {
minCount = results[i];
}
}
for (int i = 0; i < results.length; i++) {
if (results[i] == minCount) {
System.out.println("Element: " + i + ", Number of occurences: "
+ minCount);
}
}
}
Output:
Element: 4, Number of occurences: 1
Element: 7, Number of occurences: 1
This version is also quite a bit cleaner and removes a bunch of unnecessary variables.
This is not as elegant as Iwburks answer, but I was just playing around with a 2D array and came up with this:
public static void main(String[] args)
{
int[] array = { 3, 3, 3, 2, 2, -4, 4, 5, 4 };
int count = 0;
int maxCount = Integer.MAX_VALUE;
int[][] results = new int[array.length][];
int k = 0; // To keep index in 'results'
for (int i = 0; i < array.length; i++) {
for (int j = 0; j < array.length; j++) {
if (array[j] == array[i]) {
count++;
}
}
if (count <= maxCount) {
maxCount = count;
results[k++] = new int[]{array[i], count};
}
count = 0; // Reset 'count'
}
// Printing result
for (int h = 0; h < results.length; h++) {
if (results[h] != null && results[h][1] == maxCount ) {
System.out.println("Element: " + results[h][0] + ", Number of occurences: " + maxCount);
}
}
Prints
Element: -4, Number of occurences: 1
Element: 5, Number of occurences: 1
In your example above, it looks like you are only using ints. I would suggest the following solution in that situation. This will find the last number in the array with the least occurrences. I assume you don't want an object-oriented approach either.
int [] array = { 5, 1, 2, 40, 2, -1, 3, 2, 5, 4, 2, 40, 2, 1, 4 };
//initialize this array to store each number and a count after it so it must be at least twice the size of the original array
int [] countArray = new int [array.length * 2];
//this placeholder is used to check off integers that have been counted already
int placeholder = Integer.MAX_VALUE;
int countArrayIndex = -2;
for(int i = 0; i < array.length; i++)
{
int currentNum = array[i];
//do not process placeholders
if(currentNum == placeholder){
continue;
}
countArrayIndex = countArrayIndex + 2;
countArray[countArrayIndex] = currentNum;
int count = 1; //we know there is at least one occurence of this number
//loop through each preceding number
for(int j = i + 1; j < array.length; j++)
{
if(currentNum == array[j])
{
count = count + 1;
//we want to make sure this number will not be counted again
array[j] = placeholder;
}
}
countArray[countArrayIndex + 1] = count;
}
//In the code below, we loop through inspecting each number and it's respected count to determine which one occurred least
//We choose Integer.MAX_VALUE because it's a number that easily indicates an error
//We did not choose -1 or 0 because these could be actual numbers in the array
int minNumber = Integer.MAX_VALUE; //actual number that occurred minimum amount of times
int minCount = Integer.MAX_VALUE; //actual amount of times the number occurred
for(int i = 0; i <= countArrayIndex; i = i + 2)
{
if(countArray[i+1] <= minCount){
minNumber = countArray[i];
minCount = countArray[i+1];
}
}
System.out.println("The number that occurred least was " + minNumber + ". It occured only " + minCount + " time(s).");